You wouldn't be able to plot that particular graph unless u define the values a, b and, c.

What you need to recognise is how the values a, b and c affects the shape of the mentioned equation. Next under what conditions would no real solution exists regardless of what b and c take (what does this mean graphically).

For the equation y=acos(bx+c), I know that a affects the amplitude and b affects the period. I don’t know about C.

As for the conditions where there can be no real solutions for acos(bx+c), I think that the inequality I gave in the post is one condition. I’m not aware of any other conditions unless it is that -1<= cosx =<1 but I don’t know how to use it in the question.

Unsure if your lecturer failed to mention this, the solutions to the equation are the x coordinates of the intersections of the equations y = (2x² + 1)/(x - 1) and y = a cos (bx + c). You get the expression in the equation when trying to solve them simultaneously.

If there are no intersections, it implies there are no real roots.

You might want to explore with the online graphing software https://www.desmos.com/calculator to recall from your o levels how the variables a, b and c affect the cosine graph.

Hi if I’m understanding you correctly.

The given equation is
2x^2+1=a(x-1)cos(bx+c)
divide (x-1)
(2x^2+1)/x-1=acos(bx+c)
My understanding is that the minimum and maximum points of (2x^2+1)/x-1 can be used to find the range of y where acos(bx+c) exists so that there are no real solutions for acos(bx+c).

Is this coherent?

Yes.

You will realise, after knowing how the cosine curve varies, that for no real roots to occur for all possible values of b and c the cosine curve needs to lie strictly between the maximum and minimum points for this question. And this is affected by a only.